A bookshelf contains 6 distinct books of Mathematics and 5 distinct books of Physics. From these 11 books, 6 books are chosen at random. Let $$X$$ be the absolute value of the difference between the number of Mathematics books chosen and the number of Physics books chosen. If $$\alpha$$ is the mean of the random variable $$X$$, then the value of $$77\alpha$$ is ___.

We are given a total of 11 books: 6 distinct Mathematics books and 5 distinct Physics books. We choose 6 books at random from this shelf.

Let $$ M $$ be the number of Mathematics books chosen and $$ P $$ be the number of Physics books chosen. Since a total of 6 books are selected:

$$ M + P = 6 \implies P = 6 - M $$

The random variable $$ X $$ represents the absolute value of their difference:

$$ X = |M - P| = |M - (6 - M)| = |2M - 6| $$

Step 1: Determine the Possible Outcomes and Counts

The total number of ways to choose 6 books out of 11 is given by the combination:

$$ \text{Total Ways} = \binom{11}{6} = \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 462 $$

Since there are only 5 Physics books available, the minimum number of Mathematics books we can choose is 1 (leaving 5 Physics books). Thus, the possible values for $$ M $$ range from 1 to 6:

Case 1: $$ M = 1, P = 5 $$

$$ X = |1 - 5| = 4 $$

$$ \text{Ways} = \binom{6}{1} \times \binom{5}{5} = 6 \times 1 = 6 $$

Case 2: $$ M = 2, P = 4 $$

$$ X = |2 - 4| = 2 $$

$$ \text{Ways} = \binom{6}{2} \times \binom{5}{4} = 15 \times 5 = 75 $$

Case 3: $$ M = 3, P = 3 $$

$$ X = |3 - 3| = 0 $$

$$ \text{Ways} = \binom{6}{3} \times \binom{5}{3} = 20 \times 10 = 200 $$

Case 4: $$ M = 4, P = 2 $$

$$ X = |4 - 2| = 2 $$

$$ \text{Ways} = \binom{6}{4} \times \binom{5}{2} = 15 \times 10 = 150 $$

Case 5: $$ M = 5, P = 1 $$

$$ X = |5 - 1| = 4 $$

$$ \text{Ways} = \binom{6}{5} \times \binom{5}{1} = 6 \times 5 = 30 $$

Case 6: $$ M = 6, P = 0 $$

$$ X = |6 - 0| = 6 $$

$$ \text{Ways} = \binom{6}{6} \times \binom{5}{0} = 1 \times 1 = 1 $$

Step 2: Calculate the Expected Value $$ \alpha $$

The mean $$ \alpha $$ of the random variable $$ X $$ is found by multiplying each value of $$ X $$ by its corresponding number of ways, summing them up, and dividing by the total ways:

$$ \alpha = \frac{\sum (X \times \text{Ways})}{\text{Total Ways}} $$

Let us calculate the numerator sum:

$$ \sum (X \times \text{Ways}) = (4 \times 6) + (2 \times 75) + (0 \times 200) + (2 \times 150) + (4 \times 30) + (6 \times 1) $$

$$ \sum (X \times \text{Ways}) = 24 + 150 + 0 + 300 + 120 + 6 = 600 $$

So, the mean is:

$$ \alpha = \frac{600}{462} $$

Step 3: Compute the Value of $$ 77\alpha $$

We multiply the mean by 77:

$$ 77\alpha = 77 \times \frac{600}{462} $$

Since $$ 462 = 77 \times 6 $$, we can simplify the expression:

$$ 77\alpha = \frac{600}{6} = 100 $$

Therefore, the final value of $$ 77\alpha $$ is 100.